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3 years ago

Light is not a wave because it does not need a medium

2 answers:

7 0

It's a tough question as to whether light is a wave or a particle..

The statement is FALSE. Not all types of waves need a medium to propagate. Light can travel without a medium.

Extra Information:
Light is a longitudinal wave, sound is transverse.
Longitudinal waves do not need a medium.
Transverse waves need a medium.
This is why, for example, in the vacuum of space, light travels from stars, but sound does not.

ss7ja [257]3 years ago

5 0

the asnwer is false........

You might be interested in

A lorry of mass 4000 kg is travelling at a speed of 4 m/s.

bulgar [2K]

Answer:

KE = 32,000J

v = 8m/s

Explanation:

KE = .5*m*v²

KE = .5*4000kg*(4m/s)²

KE = 32,000J

32,000J = .5*1000kg*v²

v² = 64

v = 8m/s

7 0

2 years ago

A hot air balloon is traveling vertically upward at a constant speed of 4.5 m/s. When it is 28 m above the ground, a package is

ella [17]

Answer:

1.97 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow 21=4.5t+\frac{1}{2}\times -9.8\times t^2\\\Rightarrow 21=-4.5t-4.9t^2\\\Rightarrow 4.9t^2+4.5t-28=0\\\Rightarrow 49t^2+45t-280=0$

Solving the above equation we get

$t=\frac{-45+\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}, \frac{-45-\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}\\\Rightarrow t=1.97, -2.89$

So, the time the package was in the air is 1.97 seconds

3 0

2 years ago

What does the speed of sound depend on?

melamori03 [73]

Assuming it's an ideal material, the answer is A.

The speed of sound depends on the medium that the sound travels through.

5 0

2 years ago

Read 2 more answers

The scientific method can easily be misinterpreted as ___________. Question 1 options: a great magical trick, or slide of hands,

evablogger [386]

Answer:

I think the answer is

a "cookbook" recipe for performing scientific investigations

Explanation:

5 0

2 years ago

Read 2 more answers

an object is placed 15.8 cm in front of a thin converging lens with an unknown focal length. if a real image forms behind the le

storchak [24]

<span>On what:

f (is the focal length of the lens) = ?
p (is the distance from the object to the lens) =15.8 cm
p' (is the distance from the image to the spherical lens) = 4.2 cm

</span><span>Using the Gaussian equation, to know where the object is situated (distance from the point).
</span>
$\frac{1}{f} = \frac{1}{p} + \frac{1}{p'}$
$\frac{1}{f} = \frac{1}{15.8} + \frac{1}{4.2}$
$\frac{1}{f} = \frac{2.1}{33.18} + \frac{7.9}{33.18}$
$\frac{1}{f} = \frac{10}{33.18}$
Product of extremes equals product of means:
$10*f = 1*33.18$
$10f = 33.18$
$f = \frac{33.18}{10}$
$\boxed{\boxed{f = 3.318\:cm}}\end{array}}\qquad\quad\checkmark$

6 0

3 years ago